Nancy plans to invest \($10,500\) into two different bonds to spread out her risk. The first bond has an annual return of \(10%\), and the second bond has an annual return of \(6%\). To receive an \(8.5%\) return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.

Finding the Inverse of a Matrix

We know that the multiplicative inverse of a real number \(a\) is \(a^{−1}\), so

\[aa^{−1}=a^{−1}a=\left(\dfrac{1}{a}\right)a=1 \label{eq0}\]

For example, consider the scalar multiplication situation

\[2^{−1}=\dfrac{1}{2} \nonumber\]

therefore from Equation \ref{eq0}

\[\left(\dfrac{1}{2}\right)2=1. \nonumber\]

The multiplicative inverse of a matrix is similar in concept, except that the product of matrix \(A\) and its inverse \(A^{−1}\) equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by \(I_n\) where \(n\) represents the dimension of the matrix. Equations \ref{eq1} and \ref{eq2} are the identity matrices for a \(2×2\) matrix and a \(3×3\) matrix, respectively:

A matrix that has a multiplicative inverse is called aninvertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility,

\[AA^{−1}=A^{−1}A=I\]

is a requirement. Not all square matrices have an inverse, but if \(A\) is invertible, then \(A^{−1}\) is unique. We will look at two methods for finding the inverse of a \(2 × 2\) matrix and a third method that can be used on both \(2 × 2\) and \(3 × 3\) matrices.

Definitions: THE IDENTITY MATRIX AND MULTIPLICATIVE INVERSE

The identity matrix, \(I_n\), is a square matrix containing ones down the main diagonal and zeros everywhere else.

Finding the Multiplicative Inverse Using Matrix Multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication.

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, \(1\). Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is \(0\).

\(1a−2c=1\space R_1\)

\(2a−3c=0\space R_2\)

Using row operations, multiply and add as follows: \((−2)R_1+R_2\rightarrow R_2\). Add the equations, and solve for \(c\).

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, \(0\). Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

\(1b−2d=0\space R_1\)

\(2b−3d=1\space R_2\)

Using row operations, multiply and add as follows: \((−2)R_1+R_2=R_2\). Add the two equations and solve for \(d\).

There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

Finding the Multiplicative Inverse of \(3×3\) Matrices

Unfortunately, we do not have a formula similar to the one for a \(2×2\) matrix to find the inverse of a \(3×3\) matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.

To begin, we write the augmented matrix with the identity on the right and \(A\) on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

How to: Given a \(3 × 3\) matrix, find the inverse

Write the original matrix augmented with the identity matrix on the right.

Use elementary row operations so that the identity appears on the left.

What is obtained on the right is the inverse of the original matrix.

Use matrix multiplication to show that \(AA^{−1}=I\) and \(A^{−1}A=I\).

Solving a System of Linear Equations Using the Inverse of a Matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: \(X\) is the matrix representing the variables of the system, and \(B\) is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as

\(AX=B\)

To solve a system of linear equations using an inverse matrix, let \(A\) be the coefficient matrix, let \(X\) be the variable matrix, and let \(B\) be the constant matrix. Thus, we want to solve a system \(AX=B\). For example, look at the following system of equations.

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, \((2^{−1}) 2=\left(\dfrac{1}{2}\right) 2=1\). To solve a single linear equation \(ax=b\) for \(x\), we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of \(a\). Thus,

The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a \(2 × 2\) system and then move on to a \(3 × 3\) system.

Note: SOLVING A SYSTEM OF EQUATIONS USING THE INVERSE OF A MATRIX

Given a system of equations, write the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\). Then

Notice in the first step we multiplied both sides of the equation by \(A^{−1}\), but the \(A^{−1}\) was to the left of \(A\) on the left side and to the left of \(B\) on the right side. Because matrix multiplication is not commutative, order matters.

Example \(\PageIndex{8}\): Solving a 3 × 3 System Using the Inverse of a Matrix

Key Concepts

An identity matrix has the property \(AI=IA=A\). See Example \(\PageIndex{1}\).

An invertible matrix has the property \(AA^{−1}=A^{−1}A=I\). See Example \(\PageIndex{2}\).

Use matrix multiplication and the identity to find the inverse of a \(2×2\) matrix. See Example \(\PageIndex{3}\).

The multiplicative inverse can be found using a formula. See Example \(\PageIndex{4}\).

Another method of finding the inverse is by augmenting with the identity. See Example \(\PageIndex{5}\).

We can augment a \(3×3\) matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example \(\PageIndex{6}\).

Write the system of equations as \(AX=B\), and multiply both sides by the inverse of \(A\): \(A^{−1}AX=A^{−1}B\). See Example \(\PageIndex{7}\) and Example \(\PageIndex{8}\).

We can also use a calculator to solve a system of equations with matrix inverses. See Example \(\PageIndex{9}\).

Contributors

Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.

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